Lösung 2.3:9b - Versionsgeschichte

Tek: Replaced figure with metapost figure

2009-08-19T13:13:38Z

Replaced figure with metapost figure

← Nächstältere Version		Version vom 13:13, 19. Aug. 2009
Zeile 19:		Zeile 19:


-	<center>~~[[Image~~:~~2_3_9b-~~2(2)~~.gif]]~~</center>	+	<center>{{:2.3.9b - Solution - The parabola y = x² - 5x + 6 and points (2,0) and (3,0)}}</center>

Markus.bez um 14:03, 9. Jun. 2009

2009-06-09T14:03:03Z

← Nächstältere Version		Version vom 14:03, 9. Jun. 2009
Zeile 1:		Zeile 1:
-	Die Schnittpunkte sind die Punkte die auf der Parabel und auf der ''x''-Achse liegen, und also die beiden Gleichungen <math>y=x^{2}-5x+6</math> und <math>y=0</math> erfüllen,	+	Die Schnittpunkte sind die Punkte, die auf der Parabel und auf der ''x''-Achse liegen, und also die beiden Gleichungen <math>y=x^{2}-5x+6</math> und <math>y=0</math> erfüllen:

	{{Abgesetzte Formel\|\|<math>\left\{\begin{align}		{{Abgesetzte Formel\|\|<math>\left\{\begin{align}
Zeile 6:		Zeile 6:
	\end{align}\right.</math>}}		\end{align}\right.</math>}}

-	Wir ~~bekommen~~ direkt dass <math>y=0</math>, und dass <math>x</math> die Gleichung <math>x^{2}-5x+6=0\,</math> erfüllt. Durch quadratische Ergänzung erhalten wir	+	Wir erhalten direkt, dass <math>y=0</math> und dass <math>x</math> die Gleichung <math>x^{2}-5x+6=0\,</math> erfüllt. Durch quadratische Ergänzung erhalten wir

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 14:		Zeile 14:
	\end{align}</math>}}		\end{align}</math>}}

-	Dies ~~gibt~~ die Lösungen <math>x=\tfrac{5}{2}\pm\tfrac{1}{2}</math>, also <math>x=\tfrac{5}{2}-\tfrac{1}{2}=\tfrac{4}{2}=2</math> und <math>x=\tfrac{5}{2}+\tfrac{1}{2}=\tfrac{6}{2}=3</math>.	+	Dies ergibt die Lösungen <math>x=\tfrac{5}{2}\pm\tfrac{1}{2}</math>, also <math>x=\tfrac{5}{2}-\tfrac{1}{2}=\tfrac{4}{2}=2</math> und <math>x=\tfrac{5}{2}+\tfrac{1}{2}=\tfrac{6}{2}=3</math>.

-	Die Schnittpunkte ~~sing~~ also (2,0) und (3,0).	+	Die Schnittpunkte sind also (2,0) und (3,0).


	<center>[[Image:2_3_9b-2(2).gif]]</center>		<center>[[Image:2_3_9b-2(2).gif]]</center>

Martin um 19:58, 16. Mär. 2009

2009-03-16T19:58:38Z

← Nächstältere Version		Version vom 19:58, 16. Mär. 2009
Zeile 1:		Zeile 1:
-	~~The points of intersection are those points on the curve which also lie on the~~ ''x''-~~axis~~, ~~i.e. they are those points which satisfy both the equation of the curve~~	+	Die Schnittpunkte sind die Punkte die auf der Parabel und auf der ''x''-Achse liegen, und also die beiden Gleichungen <math>y=x^{2}-5x+6</math> und <math>y=0</math> erfüllen,
-	<math>y=x^{2}-5x+6</math> ~~and the equation of the ''x''-axis~~ <math>y=0</math>,	+

	{{Abgesetzte Formel\|\|<math>\left\{\begin{align}		{{Abgesetzte Formel\|\|<math>\left\{\begin{align}
Zeile 7:		Zeile 6:
	\end{align}\right.</math>}}		\end{align}\right.</math>}}

-	~~This system of equations gives directly that~~ <math>y=0</math> ~~and that~~ <math>x</math> ~~must satisfy the quadratic equation~~ <math>x^{2}-5x+6=0\,</math>. ~~By completing the square, we obtain that the left-hand side is~~	+	Wir bekommen direkt dass <math>y=0</math>, und dass <math>x</math> die Gleichung <math>x^{2}-5x+6=0\,</math> erfüllt. Durch quadratische Ergänzung erhalten wir

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 15:		Zeile 14:
	\end{align}</math>}}		\end{align}</math>}}

-	~~and this gives that the equation has solutions~~ <math>x=\tfrac{5}{2}\pm\tfrac{1}{2}</math>, ~~i.e.~~ <math>x=\tfrac{5}{2}-\tfrac{1}{2}=\tfrac{4}{2}=2</math> ~~and~~ <math>x=\tfrac{5}{2}+\tfrac{1}{2}=\tfrac{6}{2}=3</math>.	+	Dies gibt die Lösungen <math>x=\tfrac{5}{2}\pm\tfrac{1}{2}</math>, also <math>x=\tfrac{5}{2}-\tfrac{1}{2}=\tfrac{4}{2}=2</math> und <math>x=\tfrac{5}{2}+\tfrac{1}{2}=\tfrac{6}{2}=3</math>.

-	~~The intersection points are therefore~~ (2,0) ~~and~~ (3,0).	+	Die Schnittpunkte sing also (2,0) und (3,0).


	<center>[[Image:2_3_9b-2(2).gif]]</center>		<center>[[Image:2_3_9b-2(2).gif]]</center>

Tekbot: hat „Solution 2.3:9b“ nach „Lösung 2.3:9b“ verschoben: Robot: moved page

2008-10-22T14:15:45Z

hat „Solution 2.3:9b“ nach „Lösung 2.3:9b“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 14:15, 22. Okt. 2008

Tekbot: Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

2008-10-22T08:35:08Z

Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

← Nächstältere Version		Version vom 08:35, 22. Okt. 2008
Zeile 2:		Zeile 2:
	<math>y=x^{2}-5x+6</math> and the equation of the ''x''-axis <math>y=0</math>,		<math>y=x^{2}-5x+6</math> and the equation of the ''x''-axis <math>y=0</math>,

-	{{~~Displayed math~~\|\|<math>\left\{\begin{align}	+	{{Abgesetzte Formel\|\|<math>\left\{\begin{align}
	y&=x^{2}-5x+6\,,\\		y&=x^{2}-5x+6\,,\\
	y&=0\,\textrm{.}		y&=0\,\textrm{.}
Zeile 9:		Zeile 9:
	This system of equations gives directly that <math>y=0</math> and that <math>x</math> must satisfy the quadratic equation <math>x^{2}-5x+6=0\,</math>. By completing the square, we obtain that the left-hand side is		This system of equations gives directly that <math>y=0</math> and that <math>x</math> must satisfy the quadratic equation <math>x^{2}-5x+6=0\,</math>. By completing the square, we obtain that the left-hand side is

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	x^{2} - 5x + 6 &= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \Bigl(\frac{5}{2}\Bigr)^{2} + 6\\[5pt]		x^{2} - 5x + 6 &= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \Bigl(\frac{5}{2}\Bigr)^{2} + 6\\[5pt]
	&= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \frac{25}{4} + \frac{24}{4}\\[5pt]		&= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \frac{25}{4} + \frac{24}{4}\\[5pt]

Tek um 14:11, 29. Sep. 2008

2008-09-29T14:11:17Z

← Nächstältere Version		Version vom 14:11, 29. Sep. 2008
Zeile 1:		Zeile 1:
-	The points of intersection are those points on the curve which also lie on the	+	The points of intersection are those points on the curve which also lie on the ''x''-axis, i.e. they are those points which satisfy both the equation of the curve
-	~~<math>~~~x~~</math>~~	+	<math>y=x^{2}-5x+6</math> and the equation of the ''x''-axis <math>y=0</math>,
-	-axis, i.e. they are those points which satisfy both the equation of the curve	+
-	<math>y=x^~~{\text~~{2}}-~~\text{5}x~~+~~\text{~~6}</math>	+
-	and the equation of the	+
-	~~<math>~~~x~~</math>~~	+
-	-axis	+
-	<math>y=0</math>,	+

		+	{{Displayed math\|\|<math>\left\{\begin{align}
		+	y&=x^{2}-5x+6\,,\\
		+	y&=0\,\textrm{.}
		+	\end{align}\right.</math>}}

-	<math>~~\left\{ \begin{matrix}~~	+	This system of equations gives directly that <math>y=0</math> and that <math>x</math> must satisfy the quadratic equation <math>x^{2}-5x+6=0\,</math>. By completing the square, we obtain that the left-hand side is
-	y=x^~~{\text~~{2}}-~~\text{5}x~~+~~\text{~~6~~} \\~~	+
-	y=0\~~quad \quad \quad \quad \\~~	+
-	~~\end{matrix} \right.~~</math>	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	x^{2} - 5x + 6 &= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \Bigl(\frac{5}{2}\Bigr)^{2} + 6\\[5pt]
		+	&= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \frac{25}{4} + \frac{24}{4}\\[5pt]
		+	&= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \frac{1}{4}
		+	\end{align}</math>}}

-	~~This system of equations~~ gives ~~directly~~ that	+	and this gives that the equation has solutions <math>x=\tfrac{5}{2}\pm\tfrac{1}{2}</math>, i.e. <math>x=\tfrac{5}{2}-\tfrac{1}{2}=\tfrac{4}{2}=2</math> and <math>x=\tfrac{5}{2}+\tfrac{1}{2}=\tfrac{6}{2}=3</math>.
-	<math>y=0</math>	+
-	~~and that~~	+
-	<math>~x</math>	+
-	~~must satisfy the second-order equation~~	+
-	<math>x^{\~~text~~{2}}-\~~text~~{5}x+\~~text~~{6}=0</math>	+
-	. ~~By completing the square, we obtain that the left-hand side is~~	+

		+	The intersection points are therefore (2,0) and (3,0).

-	<math>\begin{align}
-	& x^{\text{2}}-\text{5}x+\text{6}=\left( x-\frac{5}{2} \right)^{2}-\left( \frac{5}{2} \right)^{2}+6 \\
-	& =\left( x-\frac{5}{2} \right)^{2}-\frac{25}{4}+\frac{24}{4}=\left( x-\frac{5}{2} \right)^{2}-\frac{1}{4} \\
-	\end{align}</math>

-		+	<center>[[Image:2_3_9b-2(2).gif]]</center>
-	~~and this gives that the equation has solutions~~	+
-		+
-	~~<math>x=\frac{5}{2}\pm \frac{1}{2}</math>, i.e.~~	+
-	~~<math>x=\frac{5}{2}-\frac{1}{2}=\frac{4}{2}=2</math>~~	+
-	~~and~~	+
-	~~<math>x=\frac{5}{2}+\frac{1}{2}=\frac{6}{2}=3</math>.~~	+
-		+
-	~~The intersection points are therefore~~	+
-	~~<math>\left( 2 \right.,\left. 0 \right)</math>~~	+
-	~~and~~	+
-	~~<math>\left( 3 \right.,\left. 0 \right)</math>.~~	+
-		+
-		+
-		+
-	~~{{NAVCONTENT_START}}~~	+
-	<center> [[Image:2_3_9b-2(2).gif]] </center>	+
-	~~{{NAVCONTENT_STOP}}~~	+

Ian um 11:57, 21. Sep. 2008

2008-09-21T11:57:08Z

← Nächstältere Version		Version vom 11:57, 21. Sep. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	The points of intersection are those points on the curve which also lie on the
-	<~~center~~> ~~[[Image:2_3_9b~~-1(2).~~gif]]~~ </~~center~~>	+	<math>~x</math>
-	{{~~NAVCONTENT_STOP~~}}	+	-axis, i.e. they are those points which satisfy both the equation of the curve
		+	<math>y=x^{\text{2}}-\text{5}x+\text{6}</math>
		+	and the equation of the
		+	<math>~x</math>
		+	-axis
		+	<math>y=0</math>,
		+
		+
		+	<math>\left\{ \begin{matrix}
		+	y=x^{\text{2}}-\text{5}x+\text{6} \\
		+	y=0\quad \quad \quad \quad \\
		+	\end{matrix} \right.</math>
		+
		+
		+	This system of equations gives directly that
		+	<math>y=0</math>
		+	and that
		+	<math>~x</math>
		+	must satisfy the second-order equation
		+	<math>x^{\text{2}}-\text{5}x+\text{6}=0</math>
		+	. By completing the square, we obtain that the left-hand side is
		+
		+
		+	<math>\begin{align}
		+	& x^{\text{2}}-\text{5}x+\text{6}=\left( x-\frac{5}{2} \right)^{2}-\left( \frac{5}{2} \right)^{2}+6 \\
		+	& =\left( x-\frac{5}{2} \right)^{2}-\frac{25}{4}+\frac{24}{4}=\left( x-\frac{5}{2} \right)^{2}-\frac{1}{4} \\
		+	\end{align}</math>
		+
		+
		+	and this gives that the equation has solutions
		+
		+	<math>x=\frac{5}{2}\pm \frac{1}{2}</math>, i.e.
		+	<math>x=\frac{5}{2}-\frac{1}{2}=\frac{4}{2}=2</math>
		+	and
		+	<math>x=\frac{5}{2}+\frac{1}{2}=\frac{6}{2}=3</math>.
		+
		+	The intersection points are therefore
		+	<math>\left( 2 \right.,\left. 0 \right)</math>
		+	and
		+	<math>\left( 3 \right.,\left. 0 \right)</math>.
		+
		+
		+
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
	<center> [[Image:2_3_9b-2(2).gif]] </center>		<center> [[Image:2_3_9b-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Tekbot: Lösning 2.3:9b moved to Solution 2.3:9b: Robot: moved page

2008-09-09T09:05:23Z

Lösning 2.3:9b moved to Solution 2.3:9b: Robot: moved page

← Nächstältere Version

Version vom 09:05, 9. Sep. 2008

Tekbot: Robot: Automated text replacement (-[[Bild: +[[Image:)

2008-08-21T06:49:49Z

Robot: Automated text replacement (-[[Bild: +[[Image:)

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-	<center> [[~~Bild~~:2_3_9b-2(2).gif]] </center>	+	<center> [[Image:2_3_9b-2(2).gif]] </center>
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2008-04-01T10:28:51Z

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